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factorial

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Source: — Data Sufficiency |

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by Rahul@gurome » Thu Jan 20, 2011 2:39 am
Ramit88 wrote:Does the integer g have a factor f such that 1<f<g ?

1. g>3!
2. 11! + 11 >= g >=11! + 2
Take a look at this post : https://www.beatthegmat.com/integers-pro ... tml#320994

The problems are almost same.
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by aleph777 » Thu Jan 20, 2011 11:40 am
Want to be sure I have a grasp on this solution. Can someone confirm?

The question is essentially asking: is G prime?

STATEMENT 1 in insufficient because: g>3! means g could be ANY number greater than 1 * 2 * 3. Therefore, g could be 8, 9, or 10, in which case it would have a smaller factor that is greater than 1. But it could also be 7, 11, etc., in which case f would not exist.

STATEMENT 2 is sufficient, however, because: g could be one of 10 different consecutive integers that are composed of 11! plus some number between 2 and 11, which means the number being added can never push the new value into a prime since it's appearing both in the factorial and as an addition, and is therefore a multiple.
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by Adam@Knewton » Thu Jan 20, 2011 2:19 pm
aleph777 wrote:Want to be sure I have a grasp on this solution. Can someone confirm?

The question is essentially asking: is G prime?

STATEMENT 1 in insufficient because: g>3! means g could be ANY number greater than 1 * 2 * 3. Therefore, g could be 8, 9, or 10, in which case it would have a smaller factor that is greater than 1. But it could also be 7, 11, etc., in which case f would not exist.

STATEMENT 2 is sufficient, however, because: g could be one of 10 different consecutive integers that are composed of 11! plus some number between 2 and 11, which means the number being added can never push the new value into a prime since it's appearing both in the factorial and as an addition, and is therefore a multiple.
Exactly, exactly right. Kudos!
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by alextheodosi » Wed Mar 06, 2013 7:20 pm
Hi,

I have just done this problem. I agree with statement 1.

Regarding statement 2:

when i look at this problem

I see that 11! is common to both sides of this inequality, therefore we can subtract 11! from both sides can't we?
And then 11>=g>=2? which presents the same as statement 1, ie includes some primes and if G can be prime than the statement is not sufficient.

While i understand the explanation below, I jumped to seeing the above (is this mathematically correct? I think it is? and if it is then why does it change the outcome of the problem?)

I don't understand why the above is not correct?

Thanks,
Alex
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